Practice Questions
Find the 5th term of the GP: 3, 6, 12, 24, ...
$a = 3$, $r = 2$. $a_5 = 3 \\times 2^{4} = 3 \\times 16 = 48$.
Find the sum of the first 4 terms of the GP: 2, 6, 18, 54, ...
$S_4 = \\frac{a(r^n - 1)}{r - 1} = \\frac{2(3^4 - 1)}{3 - 1} = \\frac{2(80)}{2} = 80$.
What is the sum of the infinite GP: 1, 1/3, 1/9, 1/27, ...?
$a = 1$, $r = 1/3$. $S_\\infty = \\frac{a}{1-r} = \\frac{1}{1-1/3} = \\frac{1}{2/3} = \\frac{3}{2}$.
The 3rd term of a GP is 12 and the 6th term is 96. What is the first term?
$a_6/a_3 = r^3 = 96/12 = 8 \\Rightarrow r = 2$. $a_3 = a \\times 4 = 12 \\Rightarrow a = 3$.
The 4th term of a GP is 24 and the 7th term is 192. Find the first term and common ratio.
$ar^3=24$, $ar^6=192$. $r^3=192/24=8 \\Rightarrow r=2$. $a(8)=24 \\Rightarrow a=3$.
Find the sum of the first 6 terms of the GP 3, 6, 12, ...
$a=3$, $r=2$. $S_6 = 3(2^6-1)/(2-1) = 3\\times63 = 189$.
If the sum to infinity of a GP is 20 and the first term is 5, find the common ratio.
$S_\\infty = a/(1-r) \\Rightarrow 20 = 5/(1-r) \\Rightarrow 1-r = 5/20 = 0.25 \\Rightarrow r = 0.75$.
Insert two geometric means between 4 and 32.
Let $4, ar, ar^2, 32$ be GP. $4r^3=32 \\Rightarrow r^3=8 \\Rightarrow r=2$. Means: $4\\times2=8$, $8\\times2=16$.
If the nth term of a GP is 128 and the first term is 2, find n (given r=2).
$ar^{n-1} = 128 \\Rightarrow 2(2)^{n-1} = 128 \\Rightarrow 2^{n-1} = 64 = 2^6 \\Rightarrow n-1=6 \\Rightarrow n=7$.
Find the product of the first 5 terms of a GP with first term 2 and ratio 3.
Product $= a^5 r^{0+1+2+3+4} = 2^5 r^{10} = 2^5\\times3^{10}$.
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